S is a set of integers such that I) If x is in S, then –x is

IMO very bad question .... they have not clearly stated the domain conditions. Kindly use the good materials only.

if such sets are considered infinite sets then A should be sufficient.

{1 -1 0 1 2 -2 }
Explanation for the above set: given 1 is in S hence -1 is also in S ( using property 1 )now -1 +1=0 is also in S by property 2 so we have 0 in S now in the set we have 1 -1 0, using property 2 if 1 and 0 are in set then 1+0 =1 is also in set hence we have 1 -1 0 1 in the set, again using property 2, if 1 and 1 are in set then 1+1=2 is also in set ,hence 2 is in set and using property 1 , if 2 is in S then -2 is also in S.

Hence if 1 is in S then -2 is also in S provided that the set is infinite

if the set can be stopped in between or if the set is a finite set then the set could be {1 -1 0 }

According to the OA the set is being considered infinite.

Why aren't the sets determined by I and II (independently) finite? That would make D the answer.

The set might be finite but that's not the point. We don't know whether -1, 0, and 1 are the only integers in the set. Please follow the links provide to understand the logic of these kind of questions better.

Responding to a pm:

I am not really sure what your query is but I think it is this: why isn't the set necessarily infinite?

You get that -1, 1 and 0 must be in the set - great. But this is not correct: "using property 2 if 1 and 0 are in set then 1+0 =1 is also in set hence we have 1 -1 0 1 in the set"

The set till now is {-1, 0, 1}
Both properties are already satisfied for each element. 1+0 = 1 is already present in the set. Why do you need to add another 1? How do you differentiate one 1 from another 1? The set could be just {-1, 0, 1} or it could have some other elements too.

The other question you quoted has a different rule. If k is in the set, k+1 must be in the set too. So the moment you put any one number in the set, all numbers after it will automatically be a part of this set.